It is not very difficult to accelerate an electron to a speed that is 99.5% of the speed of light, because it has such a very small mass. what is the ratio of the kinetic energy k to the rest energy mc2 in this case? in the definition of what we mean by kinetic energy (k = e - mc2), you must use the full relativistic formula for e, because v/c is not small compared to 1.

Question

Wesley

Physics

7 May, 04:54

It is not very difficult to accelerate an electron to a speed that is 99.5% of the speed of light, because it has such a very small mass. what is the ratio of the kinetic energy k to the rest energy mc2 in this case? in the definition of what we mean by kinetic energy (k = e - mc2), you must use the full relativistic formula for e, because v/c is not small compared to 1.

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Addisyn Pittman · Answer 1

m = m_o / √ (1-V^2/c^2) = 0.995 c

K / (m_o c^2) = ((m_o/√ (1-v^2/c^2)) v^2) / (1/2 m_o c^2) = (m_o/√ (1-v^2/c^2) (0.995) ^2 c^2) / (1/2 m_o c^2) = (2 x (0.995) ^2) / √ (1 - (0.995c) ^2/c^2) = 198.5

k/m_oc^2 = 198.5